A supply of 400V is connected across capacitors of 3μf...
A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
A.
8 x 10−4C
B.
4 x 10−2C
C.
8 x 10−3C
D.
4 x 10−8C
Correct answer is A
| CT | = | C1 × C2
C1 + C2 |
| = | 3 × 6
3 + 6 |
= 189 = 2μf
Q = CV
⇒ 2 × 10−6 × 400
⇒ 800 × 10−6C = 8 × 10−4C